find inverse Laplace transform of 1/(s^2+2s-3)

Question : find inverse Laplace transform of 1/(s^2+2s-3)

Solution:

To find inverse Laplace transform of 1/(s^2+2s-3)

Let

it can be written as

\[F\left(s\right)=\frac{1}{(s-1)(s+3)}\]

and therefore

\[F\left(s\right)=\left[\frac{1}{\left(s-1\right)}-\frac{1}{\left(s+3\right)}\right]\]

Now apply Inverse Laplace transform both sides

\[L^{-1}\left[F\left(s\right)\right]=L^{-1}\left[\frac{1}{(s-1)(s+3)}\right]\]

\[\f(t) =e^t-e^{-3t}\]

is the required inverse Laplace transform of 1/(s^2+2s-3)